Question

An $LCR$ circuit is equivalent to a damped pendulum. In an $LCR$ circuit the capacitor is charged to $Q _{0}$ and then connected to the $L$ and R as shown below : R If a student plots graphs of the square of maximum charge $\left( Q _{\max }^{2}\right)$ on the capacitor with time $(t)$ for two different values $L_{1}$ and $L_{2}$ $\left(L_{1}>L_{2}\right)$ of $L$ then which of the following represents this graph correctly ? (Plots are shematic and not drawn to scale)

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

Answer 2

$IR + L \frac{ dI }{ dt }-\frac{ q }{ C }=0$
$L \frac{ d ^{2} q }{ dt ^{2}}=- R \frac{ dq }{ dt }+\frac{ q }{ C }$
comparing with equation of damped oscillation
$d \frac{d^{2} y}{d t^{2}}=-\gamma \frac{d y}{d t}-k y$
The eqution of amplitude is $y = Ae ^{- bt }$
where $b =\frac{\gamma}{2 m }=\frac{ R }{2 L }$
$\therefore q _{\max }= q _{0} e ^{-\frac{ Rt }{2 L}}$
$\therefore q _{\max }^{2}= q _{0}^{2} e ^{-\frac{ Rt }{ L }}$
$\therefore$ time constant $\tau=\frac{ R }{ L }$
since $L_{1} > L_{2}$
$\tau_{1} < \tau_{2}$
Hence correct graph is $3 .$

The value of $Q_{\max }$ reduces because of energy dissipation in resistor. As the value of inductor increases the time taken for capacity to discharge or charge increases therefore heat dissipation time decreases. Hence corrcect graph is 3 .